Let the line L pass through (1, 1, 1) and intersect the lines (x-1)/2=(y+1)/3=(z-1)/4 and (x-3)/1=(y-4)/2=z/1. Then, which of the following points lies on the line L?

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #26

Let the line L pass through

(1, 1, 1)

and intersect the lines

\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}

and

\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}

. Then, which of the following points lies on the line L?

Clue 1 of 3: Line Intersection and Collinearity

The problem describes a line $L$ passing through a point $P$ and intersecting two other lines $L_{1}$ and $L_{2}$ . This implies that $P$ and the intersection points on $L_{1}$ and $L_{2}$ are collinear.

🧠 Full Solution Path

Step 1: Parametric Representation of Intersection Points✦ Active

Let the given point be $P (1, 1, 1)$ . Let line $L$ intersect the first line $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ$ at point $A$ . Thus, $A (2 λ + 1, 3 λ - 1, 4 λ + 1)$ . Let line $L$ intersect the second line $L_{2} : \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1} = μ$ at point $B$ . Thus, $B (μ + 3, 2 μ + 4, μ)$ .

Step 2: Apply Collinearity Condition to Find Parameters○ Expand

Since $P, A, B$ are collinear, the vectors $\vec{P A}$ and $\vec{P B}$ must be parallel. Calculate the vectors:

\vec{P A} = (2 λ, 3 λ - 2, 4 λ)

\vec{P B} = (μ + 2, 2 μ + 3, μ - 1)

For parallelism, their components are proportional:

\frac{2 λ}{μ + 2} = \frac{3 λ - 2}{2 μ + 3} = \frac{4 λ}{μ - 1}

Assuming $λ \neq 0$ (if $λ = 0$ , $A = (1, - 1, 1)$ , line $L$ is $x = 1, z = 1$ , which does not intersect $L_{2}$ ), from $\frac{2 λ}{μ + 2} = \frac{4 λ}{μ - 1}$ , we get $\frac{1}{μ + 2} = \frac{2}{μ - 1}$ . Solving this gives $μ - 1 = 2 μ + 4 \Rightarrow μ = - 5$ . Substitute $μ = - 5$ into the first two ratios: $\frac{2 λ}{- 5 + 2} = \frac{3 λ - 2}{2 (- 5) + 3} \Rightarrow \frac{2 λ}{- 3} = \frac{3 λ - 2}{- 7}$ . Solving this gives $- 14 λ = - 9 λ + 6 \Rightarrow - 5 λ = 6 \Rightarrow λ = - \frac{6}{5}$ .

💡 Teacher's Secret Hint

Remember to check for the case where $λ = 0$ or $μ = 0$ if you divide by them.

Step 3: Equation of Line L and Option Verification○ Expand

Using $P (1, 1, 1)$ and $λ = - \frac{6}{5}$ , we find point $A$ :

A = (2 (- \frac{6}{5}) + 1, 3 (- \frac{6}{5}) - 1, 4 (- \frac{6}{5}) + 1) = (- \frac{7}{5}, - \frac{23}{5}, - \frac{19}{5})

The direction vector of line $L$ is $\vec{P A} = (- \frac{7}{5} - 1, - \frac{23}{5} - 1, - \frac{19}{5} - 1) = (- \frac{12}{5}, - \frac{28}{5}, - \frac{24}{5})$ . A simpler direction vector is $(3, 7, 6)$ (by multiplying by $- \frac{5}{4}$ ). The equation of line $L$ is:

\frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}

Now, check the given options:

1. For $(4, 22, 7)$ : $\frac{4 - 1}{3} = 1$ , $\frac{22 - 1}{7} = 3$ . Not on the line.

2. For $(7, 15, 13)$ : $\frac{7 - 1}{3} = 2$ , $\frac{15 - 1}{7} = 2$ , $\frac{13 - 1}{6} = 2$ . All ratios are equal, so this point lies on line $L$ .

3. For $(10, - 29, - 50)$ : $\frac{10 - 1}{3} = 3$ , $\frac{- 29 - 1}{7} = - \frac{30}{7}$ . Not on the line.

4. For $(5, 4, 3)$ : $\frac{5 - 1}{3} = \frac{4}{3}$ , $\frac{4 - 1}{7} = \frac{3}{7}$ . Not on the line.

💡 Teacher's Secret Hint

Always verify your calculated parameters by substituting them back into all proportionality equations.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Line Intersection and Collinearity

🧠 Full Solution Path